A point at which a measurable set occupies an asymptotically full fraction of small balls, as expressed above with Lebesgue measure. The Lebesgue density theorem says that almost every point of a measurable set is a density point. On the real line, the same limiting conclusion holds for arbitrary shrinking intervals containing that point, since each interval lies in a centered interval of at most twice its length.

For any circle null set and any positive sequence ${ϵ}_j$, take infinitely many countable covers with summable total lengths. Regard their arcs as a two-index array. Choose increasing finite-square cutoffs so the length sum outside each square is smaller than the next desired bound. The finite square annuli then give finite arc families with the stated small lengths. Every point belongs to some arc in every row, hence to infinitely many annular families. This handles dense and nonclosed null sets without claiming that they have finite small covers.

Let $a_n=2^{-2^n}$ and, in angular coordinates near zero, take a bounded indicator alternating between one and zero on the rings $a_{n+1}<|t|\le a_n$. Since $a_{n+1}/a_n\to 0$, the mean over $[-a_n,a_n]$ approaches the value of the outermost ring. Its even and odd subsequences approach one and zero. Thus an integrable function can fail to have a limiting centered average at a specified point, although Lebesgue differentiation theorem ensures convergence almost everywhere.

For a finite positive measure on $[0,1)$, average its mass over each half-open dyadic interval. The resulting nonnegative martingale has constant expectation $\mu ([0,1))$: a parent interval's mass is the sum of its children's masses. Its almost-sure limit is integrable but the martingale need not be uniformly integrable. The limit recovers the density of the absolutely continuous component in the Lebesgue decomposition theorem.

For bids $1,2$ and $V>2$, write $p_i$ for player $i$'s probability of bidding $1$. The two expected payoffs are $(V-1)(1-p_j)(1-p_k)$ and $(V-2)p_j{p}_k$. Their unique symmetric equilibrium uses the displayed probability. Every permutation of $(1,0,t)$, $0\le t\le 1$, is also a Nash equilibrium: the mixed bidder is indifferent, while the two pure bidders weakly prefer their prescribed bids. There are no further equilibria. Three interior probabilities must all coincide, since subtracting two indifference equations gives a nonzero positive factor times their difference. Two interior probabilities with a pure opponent are impossible, and a single interior probability requires the other two bids to differ. Thus the nonsymmetric equilibria form six line segments with six pure endpoints, and their number is uncountable.

In the leading interaction of two sedimenting spheres, put $c=3D/4$ and let $\theta$ be the angle from the upward vertical. The equations are $\dot{r}=-(\delta +2c/r)\cos \theta$ and $r\dot{\theta }=(\delta +c/r)\sin \theta$. Their ratio separates toso $r(\delta r+2c){\sin }^2\theta$ is invariant. When $\delta ,c$ have the same nonzero sign, choose labels so both are positive. For a noncollinear encounter the invariant is positive, ensuring a strictly positive closest separation. Meanwhile $\dot{z}=-\delta -c(1+{\cos }^2\theta )/r<0$, so the vertical order reverses and eventually $r\ge |z|\to \infty$. Exactly collinear contact is outside this passing argument.

If $\delta D<0$, the leading interaction of two sedimenting spheres has a formal vertical equilibrium at $r_{\ast }=-3D/(2\delta )$. Choose labels with $\delta >0,D<0$ and put sphere one above sphere two. On that invariant vertical line, $\dot{r}=-\delta +\delta r_{\ast }/r$, so both sides approach $r_{\ast }$ and the radial eigenvalue is $-\delta /r_{\ast }$. A small tilt instead grows at rate $\delta /(2r_{\ast })$: the full configuration-space equilibrium is a saddle, not an attracting node. A physically valid far-field bound pair requires $r_{\ast }\gg a_1+a_2$, or $|D|\gg (a_1+a_2)|\delta |$. Reversing both driving forces reverses the vector field, interchanging radial attraction and repulsion, consistent with kinematic reversibility of Stokes flow.

For integrable random variable $X$ and nested sub-sigma-algebras $\mathcal{H}\subseteq \mathcal{G}$, the tower property saysThe finer-information conditional mean can be averaged within each coarser group, recovering the coarser-information conditional mean. Conversely, conditioning an already $\mathcal{H}$-measurable conditional mean on $\mathcal{G}$ leaves it unchanged. Both orders therefore recover $\mathbb{E}[X\mid \mathcal{H}]$ when the information is nested.^[ref]

For the finite example in law of total expectation, exact knowledge of $X$ gives the finer mean $X$ itself. Averaging within the groups $A,B$ recovers two and eight. Forgetting the group then gives $13/2$. The nesting condition is essential; iterated conditional expectation over nonnested sigma-algebras describes why arbitrary successive conditioning cannot simply be replaced by conditioning on an intersection.

A group of order automorphisms of a totally ordered set that is closed under pointwise maximum and minimum. Products act on the right when writing $\alpha f$; then $\alpha (fg)=(\alpha f)g$. The pointwise lattice operations make it a lattice-ordered group.

A commutative lattice-ordered group, usually written additively. Examples include real-valued function groups with pointwise addition, maximum and minimum. Such a group need not have a total order.

A group with a two-sided invariant total order. Thus it is a lattice-ordered group whose join and meet choose the larger and smaller element. An order on only one side is not sufficient here. Every linearly ordered group is torsion-free: a positive element has strictly increasing positive powers, and a negative element has a positive inverse.

A lattice-ordered group is normal-valued when every value in a lattice-ordered group is a normal subgroup of its cover. Normality in the cover is weaker than normality in the entire group. Every linearly ordered group is normal-valued.

A convex subgroup of an ordered group that is also closed under join and meet. A normal convex lattice subgroup is the kernel of the quotient lattice-ordered group homomorphism.

The displayed join is nonnegative in a lattice-ordered group. In an additive Abelian group it becomes $|g|=g\vee (-g)$. This order-theoretic absolute value need not be a real scalar.

A group homomorphism preserving both join and meet. A bijective such map is an isomorphism of lattice-ordered groups; an injective one is an embedding.

For $a,b\ge 1$, take $a^{\prime }=x\wedge a$ and $b^{\prime }=(a^{\prime }{)}^{-1}x$. Then $a^{\prime }\ge 1$ and $b^{\prime }=1\vee a^{-1}x$ is between one and $b$. Thus $x=a^{\prime }{b}^{\prime }$. Repetition splits a positive element bounded by a finite product into factors bounded by the corresponding factors. Every resulting positive factor is also bounded by $x$.

Equal sums give equal Klein j-invariants, hence the complex lattices are homothetic. If both sums are nonzero, the scaling factor obeys $a^4=a^6=1$ and is $\pm 1$. If only $G_6$ vanishes, the lattice is homothetic to the square lattice and the possible fourth roots of unity already preserve it. If only $G_4$ vanishes, it is homothetic to the hexagonal lattice and the possible sixth roots of unity preserve it. The two sums cannot vanish together because the elliptic discriminant is nonzero.

A homomorphism between lattices preserves binary joins and meets: $f(x\vee y)=f(x)\vee f(y)$ and $f(x\wedge y)=f(x)\wedge f(y)$. It is order preserving, since $x\le y$ implies $x=x\wedge y$ and hence $f(x)=f(x)\wedge f(y)$. For a map into the two-element lattice, the inverse image of zero is a lattice ideal, and meet preservation says this ideal is prime: if a meet belongs to it, at least one factor does. Conversely the indicator of the complement of a prime lattice ideal preserves both operations. Preservation of distinguished top and bottom elements is an additional requirement for homomorphisms of bounded lattices.

In a lattice, joins and meets are the pairwise least upper bounds and greatest lower bounds.

Latin squares on the same array are mutually orthogonal when every ordered pair of symbols from any two squares occurs exactly once on superposition. Pairwise balance makes the centered symbol-indicator spaces orthogonal in an additive linear regression model.